Factoring

[broken sphinx]

I am trying to factor -x^3 + 5x^2 - 8x +4 by grouping but can't seem to get the correct final form. Can someone please help?
-x^3 + 5x^2 - 8x +4

• ( x^3 - 5x^2 + 8x - 4)
• [x^2 (x-5) + 4 (2x - 1)]
  And from here, i'm stuck because I don't have the same expression in each bracket, so factoring by grouping didn't work here

[near gyro]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with: ```diff
   +close

[urban magnet]

@broken sphinx An important theorem for polynomials with integer coefficients "rational root theorem" states that the polynomial has a root x=p/q where p is a factor of the last coefficient and q is a factor of the first coeficient

In this case +4 is the last coefficient and -1 is the first coeficient so we have +4/(-1)= -4.

The possible roots of the polynomial are the factors of -4: ±1, ±2,±4.

Check that 1 is a root, 2 is also a root

So -x³+5x²-8x+4=(x-1)(x-2)(ax+b)

Can you find the a,b by expanding the right hand side and setting that the coefficients between the two sides are equal?

You will find that a=-1 and b=2 so the answer is -x³+5x²-8x+4=-(x-1)(x-2)² , you see that 2 is a double root.

After you find 1 is a root you could instead say -x³+5x²-8x+4=(x-1)(ax²+bx+c), expand the right hand side, set the corresponding coefficients between the two sides equal and find that a=-1,b=4,c=-4, and then proceed by factoring ax²+bx+c.

[broken sphinx]

What do u mean by this?

[urban magnet]

@broken sphinx Like for example if you have two equal polynomials ax³+bx²+cx+d=ex³+fx²+gx+h then the corresponding coeficients are equal a=e, b=f , c=g, h=d

-x³+5x²-8x+4=(x-1)(x-2)(ax+b)=(x²-3x+2)(ax+b)=ax³+(b-3a)x²+(3b+2a)x+2b